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What weight of Na(2)CO(3) of 95% purity ...

What weight of `Na_(2)CO_(3) of 95%` purity would be required to neutralize `45.6 mL` of `0.235 N` acid?

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The correct Answer is:
`0.5978g`;
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What weight of Na_(2)CO_(3) of 93% purity would be required to neutralise 45.6 mL of 0.235 N acid?

The concentration of solutions can be expressed in number of ways such that Normality, Molarity, Molality, Mole fractions, Strength, % by weight, % by volume and % by strength. The molarity of ionic compound is usually expressed as formality beacuse we use formula weight of ionic compound. Addition of water to a solution changes all these terms, however increase in temperature does not change molality, mole fraction and % by weight terms. The weight of Na_(2)CO_(3) sample of 95% purity required to neutralise 45.6 mL of 0.235 N acid is:

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P BAHADUR-MOLE AND EQUIVALENT CONCEPT-Exercise 9 Advanced numerical problems
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